Standard entropies of X2- Y2 and XY3 are 60- 40 and 50 JK-1mol-1 respectively- For the reaction- 12 X2 - 32Y2 - XY3- H - -30 kJ-molto be at spontaneous- the temperature should be-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Gibbs Free Energy & Spontaneity

Standard entr...

Question

Standard entropies of $X_{2}, Y_{2} a n d X Y_{3}$ are 60, 40 and $50 J K^{- 1} m o l^{- 1}$ respectively. For the reaction:
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}; △ H = - 30 k J / m o l$
to be at spontaneous, the temperature should be:

750 K

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1000 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1250 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

500 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 750 K
the reaction:
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}; △ H = - 30 k J / m o l$
$△ S_{X_{2}} = 60 J K^{- 1} m o l^{- 1}$
$△ S_{Y_{2}} = 40 J K^{- 1} m o l^{- 1}$
$△ S_{X Y_{3}} = 50 J K^{- 1} m o l^{- 1}$
$△ S = \sum S_{P r o d u c t s} - \sum S_{R e a c t a n t s}$
$△ S = △ S_{X Y_{3}} - [△ S_{X_{2}} + △ S_{Y_{2}}]$
$= 50 - [\frac{1}{2} \times 60 + \frac{3}{2} \times 40]$
$= 50 - (30 + 60) = - 40 J K^{- 1} m o l^{- 1}$

for reaction to be spontaneous putting $△ G < 0$
so if, $△ G = △ H - T △ S < 0$
i.e., $△ H < T △ S$
As both have negative sign so inequality will change.

$T < \frac{△ H}{△ S} T < \frac{- 30 \times 1000}{- 40} T < 750 K$
below this temperature the reaction will spontaneous, because $△ G < 0$

Theory:

For spontaneous process, $Δ S_{u n i v}$ is greater than 0 and minimum value of T is 0, T will be positive.

So for spontaneity $Δ G_{s y s}$ must be negative.

For Spontaneous reaction $G_{s y s}$ is less than 0
For non spontaneous reaction $G_{s y s}$ is greater than 0
For equilibrium condition $G_{s y s} = 0$

Suggest Corrections

Similar questions

Q. Standard entropies of

X_{2}, Y_{2} a n d X Y_{3}

are 60, 40 and

50 J K^{- 1} m o l^{- 1}

respectively. For the reaction:

\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}; △ H = - 30 k J / m o l

to be at spontaneous, the temperature should be:

Q. Standard entropies of

X_{2}, Y_{2}

and

X Y_{3}

, are 60, 40 and 50

K^{- 1} J m o l^{- 1}

respectively. For the reaction:

\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}, Δ H = - 30 k J

to be at equilibrium, the temperature should be:

Standard entropy of $X_{2}$ , $Y_{2}$ and $X Y_{3}$ are 60, 40 and 50 $J K^{- 1} m o l^{- 1}$ respectively. For the reaction: $\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to X Y_{3}, Δ H = - 30 k J$ to be at equilibrium, the temperature will be :

Standard entropies of

X_{2}

Y_{2}

and

{X Y}_{3}

are 60, 40 and

{50 J K}^{- 1} {m o l}^{- 1}

respectively. For the reaction

\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ {X Y}_{3}, △ H = - 30 k J

at equilibrium, the temperature should be:

Q. Standard entropy of

X_{2}, Y_{2}

and

X Y_{3}

are 60, 40 and 50

Z J K^{- 1} m o l^{- 1}

,respectively. For the reaction,

\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to X Y_{3}

Δ H = - 30 k J

, to be at equilibrium, the temperature will be :

Join BYJU'S Learning Program

Explore more

Gibbs Free Energy & Spontaneity

Standard XII Chemistry

Join BYJU'S Learning Program

Standard entropies of X2,Y2 and XY3 are 60, 40 and 50 JK−1mol−1 respectively. For the reaction: 12X2+32Y2⇌XY3; △H=−30 kJ/mol to be at spontaneous, the temperature should be:

Standard entropies of $X_{2}, Y_{2} a n d X Y_{3}$ are 60, 40 and $50 J K^{- 1} m o l^{- 1}$ respectively. For the reaction:
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}; △ H = - 30 k J / m o l$
to be at spontaneous, the temperature should be: